To prove this lemma, we use the algorithm described by [1]}. Consider the following diagram (see Figure REF ):
<FIGURE>By [2]}, Lemma REF  and [3]} we have
a FSU of \(L_1\)  is given by \(\lbrace \varepsilon _{2}, \varepsilon _{p},\sqrt{\varepsilon _{2}\varepsilon _{p}\varepsilon _{2p}}\rbrace \) , a FSU of \(L_2\)  is given by \(\lbrace \varepsilon _{2}, \sqrt{\varepsilon _{q}}, \sqrt{\varepsilon _{2q}}\rbrace \)  and a FSU of \(L_3\)  is given by \(\lbrace  \varepsilon _{2}, \varepsilon _{pq},\sqrt{\varepsilon _{pq}\varepsilon _{2pq}}\rbrace \) . It follows that,
\(E_{L_1}E_{L_2}E_{L_3}=\langle -1, \varepsilon _{2}, \varepsilon _{p}, \varepsilon _{pq}, \sqrt{\varepsilon _{q}}, \sqrt{\varepsilon _{2q}}, \sqrt{\varepsilon _{pq}\varepsilon _{2pq}}, \sqrt{\varepsilon _{2}\varepsilon _{p}\varepsilon _{2p}}\rangle .\) 
Note that a FSU of \(\mathbb {K}\)  consists of seven units chosen from those of \(L_1\) , \(L_2\)  and \(L_3\) , and from the square roots of the units of \(E_{L_1}E_{L_2}E_{L_3}\)  which are squares in \(\mathbb {K}\)  (cf. [1]}).
Thus we shall determine elements of \(E_{L_1}E_{L_2}E_{L_3}\)  which are squares in \(\mathbb {K}^+\) . Suppose that \(X\)  is an element of \(\mathbb {K}^+\)  which is a square of an element of \(E_{L_1}E_{L_2}E_{L_3}\) , so
\(X^2=\varepsilon _{2}^a\varepsilon _{p}^b \varepsilon _{pq}^c\sqrt{\varepsilon _{q}}^d\sqrt{\varepsilon _{2q}}^e\sqrt{\varepsilon _{pq}\varepsilon _{2pq}}^f\sqrt{\varepsilon _{2}\varepsilon _{p}\varepsilon _{2p}}^g, \) 
where \(a, b, c, d, e, f\)  and \(g\)  are in \(\lbrace 0, 1\rbrace \) .
We shall use norm maps from \(\mathbb {K}^+\)  to its sub-extensions to eliminate the cases of \(X^2\)  which do not occur. Set
\(G=\mathrm {Gal}(\mathbb {K}^+/\mathbb {Q})=\langle \sigma _1, \sigma _2, \sigma _3\rangle \) ,
where
\(\sigma _1(\sqrt{2})=-\sqrt{2}\) , \(\sigma _2(\sqrt{p})=-\sqrt{p}\)  and \(\sigma _3(\sqrt{q})=-\sqrt{q}\) ,
and
\(\sigma _i(\sqrt{2})=\sqrt{2}\)  for \(i\in \lbrace 2, 3\rbrace \) ,
\(\sigma _i(\sqrt{p})=\sqrt{p}\)  for \(i\in \lbrace 1, 3\rbrace \)  and
\(\sigma _i(\sqrt{q})=\sqrt{q}\)  for \(i\in \lbrace 1, 2\rbrace \) .
Hence \(L_1\) , \(L_2\)  and \(L_3\)  are
the fixed fields of the subgroups of \(G\)  generated respectively by \(\sigma _3\) , \(\sigma _2\)  and \(\sigma _2\sigma _3\) .
Let us firstly do some computations that will help in the computations of these norm maps which we shall use now and for the proof of the next lemmas as well.
Let \(L_4=\mathbb {Q}(\sqrt{p}, \sqrt{q})\)  and \(L_5=\mathbb {Q}(\sqrt{p}, \sqrt{2q})\) . By Lemma REF ,
a FSU of \(L_4\)  is \(\lbrace \varepsilon _{p}, \varepsilon _{q}, \sqrt{\varepsilon _{q}\varepsilon _{pq}}\rbrace \) ,
and a FSU of \(L_5\)  is \(\lbrace \varepsilon _{p}, \varepsilon _{2q}, \sqrt{\varepsilon _{2pq}}\rbrace \) .
Let \(u, v, t, k\)  and \( r\in \lbrace 0, 1\rbrace \) . Table REF  (page REF ) will be used to compute norm maps. Its 8-th line, for example, is constructed as follows (the other lines are constructed in the same manner). By Lemma REF , there are two integers \(y_1\)  and \(y_2\)  such that
\(\sqrt{\varepsilon _{2pq}}=\frac{\sqrt{2}}{2}(y_1\sqrt{2p}+y_2\sqrt{q})\) . Which implies that
\(\begin{array}{ll}\sqrt{\varepsilon _{2pq}}^{\sigma _1}&=\frac{-\sqrt{2}}{2}(-y_1\sqrt{2p}+y_2\sqrt{q})\\&=\frac{\frac{-\sqrt{2}}{2}(-y_1\sqrt{2p}+y_2\sqrt{q})(y_1\sqrt{2p}+y_2\sqrt{q})}{(y_1\sqrt{2p}+y_2\sqrt{q})}\\&=\frac{\frac{-\sqrt{2}}{2}(-y_1^22p+y_2^2q)}{\sqrt{2}\sqrt{\varepsilon _{2pq}}}\\&=\frac{\frac{-\sqrt{2}}{2}\cdot 2}{\sqrt{2}\sqrt{\varepsilon _{2pq}}}=\frac{-1}{\sqrt{\varepsilon _{2pq}}}\cdot \end{array}\) 
\(\begin{array}{ll}\sqrt{\varepsilon _{2pq}}^{\sigma _2}&=\frac{\sqrt{2}}{2}(-y_1\sqrt{2p}+y_2\sqrt{q})\\&=-\frac{-\sqrt{2}}{2}(-y_1\sqrt{2p}+y_2\sqrt{q})\\&=-\frac{-1}{\sqrt{\varepsilon _{2pq}}}=\frac{1}{\sqrt{\varepsilon _{2pq}}}\cdot \end{array}\) 
\(\begin{array}{ll}\sqrt{\varepsilon _{2pq}}^{\sigma _3}&=\frac{\sqrt{2}}{2}(y_1\sqrt{2p}-y_2\sqrt{q})\\&=\frac{-\sqrt{2}}{2}(-y_1\sqrt{2p}+y_2\sqrt{q})\\&=\frac{-1}{\sqrt{\varepsilon _{2pq}}}\cdot \end{array}\)
\(\begin{array}{ll}\sqrt{\varepsilon _{2pq}}^{1+\sigma _1}&=\sqrt{\varepsilon _{2pq}}\cdot \frac{-1}{\sqrt{\varepsilon _{2pq}}}=-1.\\ \sqrt{\varepsilon _{2pq}}^{1+\sigma _2}&=\sqrt{\varepsilon _{2pq}}\cdot \frac{1}{\sqrt{\varepsilon _{2pq}}}=1.\\ \sqrt{\varepsilon _{2pq}}^{1+\sigma _3}&=\sqrt{\varepsilon _{2pq}}\cdot \frac{-1}{\sqrt{\varepsilon _{2pq}}}=-1.\\\sqrt{\varepsilon _{2pq}}^{1+\sigma _1\sigma _2}&=\sqrt{\varepsilon _{2pq}}\cdot (\frac{1}{\sqrt{\varepsilon _{2pq}}})^{\sigma _{1}}=-\varepsilon _{2pq}.\\\sqrt{\varepsilon _{2pq}}^{1+\sigma _1\sigma _3}&=\sqrt{\varepsilon _{2pq}}\cdot (\frac{-1}{\sqrt{\varepsilon _{2pq}}})^{\sigma _{1}}=\varepsilon _{2pq}.\\\sqrt{\varepsilon _{2pq}}^{1+\sigma _2\sigma _3}&=\sqrt{\varepsilon _{2pq}}\cdot (\frac{-1}{\sqrt{\varepsilon _{2pq}}})^{\sigma _{2}}=-\varepsilon _{2pq}.\end{array}\) 
Now we return back to our square
\(X^2=\varepsilon _{2}^a\varepsilon _{p}^b \varepsilon _{pq}^c\sqrt{\varepsilon _{q}}^d\sqrt{\varepsilon _{2q}}^e\sqrt{\varepsilon _{pq}\varepsilon _{2pq}}^f\sqrt{\varepsilon _{2}\varepsilon _{p}\varepsilon _{2p}}^g\) ,
by applying the norm \(N_{\mathbb {K}^+/L_2}=1+\sigma _2\)  (see Table REF  page REF ) we get:
\(N_{\mathbb {K}^+/L_2}(X^2)&=&\varepsilon _{2}^{2a}\cdot (-1)^{b}\cdot 1\cdot \varepsilon _q^d \cdot \varepsilon _{2q}^e \cdot 1 \cdot (-1)^{ vg} \cdot \varepsilon _{2}^g \\&=&\varepsilon _{2}^{2a}\varepsilon _{q}^{d}\varepsilon _{2q}^{e}(-1)^{b+vg}\varepsilon _{2}^{g}.\) 
As \(\varepsilon _{q}\) , \(\varepsilon _{2q}\)  are squares in \(L_2\)  and \(N_{\mathbb {K}^+/L_2}(X^2)>0\) , so \(b+vg\equiv 0\pmod {2}\)  and \(\varepsilon _{2}^g\)  is a square in \(L_2\) . But \(\varepsilon _{2}\)  is not a square in \(L_2\) , then \(g=0\)  and thus \(b=0\) . Therefore, \(X^2\)  become
\(X^2=\varepsilon _{2}^a \varepsilon _{pq}^c\sqrt{\varepsilon _{q}}^d\sqrt{\varepsilon _{2q}}^e\sqrt{\varepsilon _{pq}\varepsilon _{2pq}}^f.\) 
Similarly, by applying \(N_{\mathbb {K}^+/L_3}=1+\sigma _2\sigma _3\)  (see Table REF  page REF ) one gets:
\(N_{\mathbb {K}^+/L_3}(X^2)=\varepsilon _{2}^{2a}\cdot \varepsilon _{pq}^{2c} \cdot (\varepsilon _{pq}\varepsilon _{2pq})^f, \) 
unfortunately, here we conclude nothing. So we will use the norm map over \(L_4=\mathbb {Q}(\sqrt{p}, \sqrt{q})\)  which is \(N_{\mathbb {K}^+/L_4}=1+\sigma _1\) .
Note that \(\lbrace \varepsilon _{p}, \varepsilon _{q}, \sqrt{\varepsilon _{q}\varepsilon _{pq}}\rbrace \)  is a FSU of \(L_4\) . Thus
\(N_{\mathbb {K}^+/L_4}(X^2)&=&(-1)^{a}\cdot \varepsilon _{pq}^{2c}\cdot (-\varepsilon _{q})^d\cdot (-1)^e \cdot (\varepsilon _{pq})^f\\&=&\varepsilon _{pq}^{2c} \cdot (-1)^{a+d+e}\cdot \varepsilon _{q}^d\cdot \varepsilon _{pq}^f>0.\) 
Since \(N_{\mathbb {K}^+/L_4}(X^2)>0\) , then \(a+d+e\equiv 0\pmod {2}\) . By Remark REF , \(2\varepsilon _{q}\)  is a square in \(\mathbb {Q}(\sqrt{q})\)  and \(2\varepsilon _{pq}\)  is a square in \(\mathbb {Q}(\sqrt{p}, \sqrt{q})\) .
So \(d=f\) , since otherwise we will get \(\varepsilon _{pq}\)  or \(\varepsilon _{q}\)  is a square in \(L_4\) . Therefore,
\(X^2=\varepsilon _{2}^a \varepsilon _{pq}^c\sqrt{\varepsilon _{q}}^d\sqrt{\varepsilon _{2q}}^e\sqrt{\varepsilon _{pq}\varepsilon _{2pq}}^d.\) 
Now we apply \(N_{\mathbb {K}^+/L_5}=1+\sigma _1\sigma _3\) , where \(L_5=\mathbb {Q}(\sqrt{p}, \sqrt{2q})\) . Note that a FSU of \(L_5\)  is given by \(\lbrace \varepsilon _{p}, \varepsilon _{2q}, \sqrt{\varepsilon _{2pq}}\rbrace \) .
We have
\(N_{\mathbb {K}^+/L_5}(X^2)&=&(-1)^{a}\cdot 1 \cdot (-1)^{d} \cdot (-\varepsilon _{2q})^{e} \cdot \varepsilon _{2pq}^d\\&=&\varepsilon _{2pq}^d \cdot (-1)^{a+d+e} \cdot \varepsilon _{2q}^e>0.\) 
So \(a+d+e\equiv 0\pmod {2}\) . As \(\varepsilon _{2q}\)  is not a square in \(L_5\) , then \(e=0\)  and \(a+d\equiv 0\pmod {2}\) . Therefore \(a=d\)  and
\(X^2=\varepsilon _{2}^d \varepsilon _{pq}^c\sqrt{\varepsilon _{q}}^d\sqrt{\varepsilon _{pq}\varepsilon _{2pq}}^d.\) 
Remark that \(\varepsilon _{pq}\)  is a square in \(\mathbb {K}^+\) , so we may put
\(X^2=\varepsilon _{2}^d\sqrt{\varepsilon _{q}}^d\sqrt{\varepsilon _{pq}\varepsilon _{2pq}}^d.\) 
Applying the norm maps from \(\mathbb {K}^+\)  to all the rest of its sub-extensions, no contradiction is obtained and we conclude nothing about \(d\) . So \(d=0\)  or 1, and thus the result.
